IOPS from RAID units

Literature from storage suppliers show performance claims of over 1000
random IOPS.
If they are really random over an array, are they not limited by the
seek time ?
So an average seek time of 8.2 ms for a SATA disk would imply a
maximum of 122 seeks per
second. So how can you claim over a 1000 random IOPS ?

wrote in message
ups.com...
Literature from storage suppliers show performance claims of over 1000
random IOPS.
If they are really random over an array, are they not limited by the seek
time ?
So an average seek time of 8.2 ms for a SATA disk would imply a maximum of
122 seeks per
second. So how can you claim over a 1000 random IOPS ?


It's worse than that for single disks. You need to add average rotational
latency. For a
7200rpm disk the disk spins (7200/60 = ) 120 times per second. One
revolution takes (1/120 = ) 8.33ms. The average rotational latency is half a
revolution, so 4.16ms. Add that to the average seek time to get 12.36 ms
average access time. That's just under 81 IOPS for a single disk. You always
have a bit of processing overhead in the electronics, so anywhere between 75
and 80 IOPS for that single disk sounds more realistic.

RAID vendors multiply the number of IOPS for each disk with the maximum
number of disks they can handle. So a disk system capable of 24 disks can
claim (24x75 = ) 1800 IOPS. Obviously if you chose to only use 5 disks or
so, you get a lot lower performance... That's the reason why often you don't
want a 2TB RAID-5 array to be built from 3 large 1TB disks, but rather use
many smaller disks.

Rob

In article . com,
wrote:

second. So how can you claim over a 1000 random IOPS ?


10 disc array = 10 seeks per 8.2ms

A raid array is capable of accessing all of the discs more or less
at once. Also, they usually cache to memory so for some data there's no seek
at all.

wrote in message
ups.com...
Literature from storage suppliers show performance claims of over 1000
random IOPS.
If they are really random over an array, are they not limited by the
seek time ?
So an average seek time of 8.2 ms for a SATA disk would imply a
maximum of 122 seeks per
second. So how can you claim over a 1000 random IOPS ?


Random IOP performance would be dependent on the size of your disk as well.

Moojit

On Wed, 18 Jul 2007 20:30:43 +0200, "Rob Turk"
wrote:

wrote in message
oups.com...
Literature from storage suppliers show performance claims of over 1000
random IOPS.
If they are really random over an array, are they not limited by the seek
time ?
So an average seek time of 8.2 ms for a SATA disk would imply a maximum of
122 seeks per
second. So how can you claim over a 1000 random IOPS ?


It's worse than that for single disks. You need to add average rotational
latency. For a
7200rpm disk the disk spins (7200/60 = ) 120 times per second. One
revolution takes (1/120 = ) 8.33ms. The average rotational latency is half a
revolution, so 4.16ms. Add that to the average seek time to get 12.36 ms
average access time. That's just under 81 IOPS for a single disk. You always
have a bit of processing overhead in the electronics, so anywhere between 75
and 80 IOPS for that single disk sounds more realistic.

RAID vendors multiply the number of IOPS for each disk with the maximum
number of disks they can handle. So a disk system capable of 24 disks can
claim (24x75 = ) 1800 IOPS. Obviously if you chose to only use 5 disks or
so, you get a lot lower performance... That's the reason why often you don't
want a 2TB RAID-5 array to be built from 3 large 1TB disks, but rather use
many smaller disks.

Rob



10K RPM FC drives get about 120ops, so 75-80 for 7200 SATA sounds
about right.

Ain't math grand?

Purely random ops is just decimating to any performance based system
unless you use something like SSD's (solid state drives, pure memory).

On the bright side, there are not many apps that are truly random.

~F

On Wed, 18 Jul 2007 20:16:00 -0500, "Moojit"
wrote:


wrote in message
oups.com...
Literature from storage suppliers show performance claims of over 1000
random IOPS.
If they are really random over an array, are they not limited by the
seek time ?
So an average seek time of 8.2 ms for a SATA disk would imply a
maximum of 122 seeks per
second. So how can you claim over a 1000 random IOPS ?


Random IOP performance would be dependent on the size of your disk as well.

Moojit



I'd be interested in if anyone concurs with that. I'm guessing that
it's based purely on seek and rotation time. I don;t think density
has anything to do with it.
I suspect your confusing larger drives with lower RPM, which would be
rotational time. So in that case yes, the ops would be less but not
due to density.

~F

wrote:
Literature from storage suppliers show performance claims of over 1000
random IOPS.
If they are really random over an array, are they not limited by the
seek time ?
So an average seek time of 8.2 ms for a SATA disk would imply a
maximum of 122 seeks per
second. So how can you claim over a 1000 random IOPS ?

Aside from the considerations already mentioned, there is the fact that
contemporary disks can reorder (most) requests in their queues to
expedite aggregate throughput. If request queues are allowed to become
fairly lengthy, this can more than double the I/O rate from what a
serial request stream would see (i.e., a disk that could service 80
small random requests per second in a serial stream could service close
to 200 if allowed to reorder a few dozen requests in its queue, and the
fastest current disks which can handle about 180 serial small random
requests per second could service close to 400 under such ideal - for
this particular benchmark - conditions).

'Short-stroking' a disk (using only a small, contiguous fraction of its
capacity to hold the relevant data) can improve performance a bit mo
current disks (very roughly - you can actually calculate this by
examining the relationship of average seek time, defined as being 1/3 of
the way across the entire disk surface, to maximum seek time from start
to end of the disk) accelerate the head for only about 1/6 of the
distance of a full-stroke seek, decelerate it for another 1/6 of the
distance at the end of the seek, and coast the rest of the way, which
just happens to minimize nominal 'average' seek times. So (neglecting
head-settling time at the end of the seek, though it rapidly becomes
non-negligible as seeks get shorter) if you limit the data to, say, 1/12
of the disk surface, then the *worst-case* seek for data on that disk
will be only half the duration of the *average* seek on a full disk (and
the *average* seek time for that data on the disk will be less than
30% of the average seek time on a full disk).

Unfortunately, you can't combine these two approaches very effectively:
short-stroking is great for serial small random requests, and queue
optimization is great for parallel small random requests (at least as
long as the long individual service times can be tolerated: that's the
price one pays for per-disk throughput in this case), but queue
optimization does nothing for serial request streams and while
short-stroking does help parallel request streams the major reduction it
makes in average seek time means that there's a great deal less that
queue optimization can accomplish (though it can still help reduce the
rotational latency component significantly).

- bill

Moojit proclaimed:

wrote in message
ups.com...

Literature from storage suppliers show performance claims of over 1000
random IOPS.
If they are really random over an array, are they not limited by the
seek time ?
So an average seek time of 8.2 ms for a SATA disk would imply a
maximum of 122 seeks per
second. So how can you claim over a 1000 random IOPS ?



Random IOP performance would be dependent on the size of your disk as well.


And how smart your raid controller and disk subsystem are at firing off
multiple I/Os in parallel whenever possible, or grouping I/Os to the
same area of disk whenever possible.

Faeandar wrote:
On Wed, 18 Jul 2007 20:16:00 -0500, "Moojit"
wrote:

wrote in message
ups.com...
Literature from storage suppliers show performance claims of over 1000
random IOPS.
If they are really random over an array, are they not limited by the
seek time ?
So an average seek time of 8.2 ms for a SATA disk would imply a
maximum of 122 seeks per
second. So how can you claim over a 1000 random IOPS ?

Random IOP performance would be dependent on the size of your disk as well.

Moojit



I'd be interested in if anyone concurs with that. I'm guessing that
it's based purely on seek and rotation time. I don;t think density
has anything to do with it.
I suspect your confusing larger drives with lower RPM, which would be
rotational time. So in that case yes, the ops would be less but not
due to density.


Random IOPS are almost totally determined by disk speed. You may see
small improvements in newer, larger, disks due to firmware changes and
cleverer read-ahead and queueing policies but nothing significant.

The read latency of ATA disks also gets bad quickly when you load them
up. FC drives handle this alot better.

Pete